数学杂志  2025, Vol. 45 Issue (3): 261-270   PDF    
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贾亿炘
韩领兄
q-Bernstein-Durrmeyer型算子在紧圆盘的复逼近等价性$(0<q<1)$
贾亿炘, 韩领兄    
内蒙古民族大学数学科学学院, 内蒙古 通辽 028000
摘要:本文研究了复q-Bernstein-Durrmeyer型算子在紧圆盘上的相关性质, 利用高阶Cauchy积分公式、泰勒展式和Bernstein不等式等方法, 获得了该算子在紧圆盘上的同时逼近和在封闭单位圆盘上的Voronovskaja型定理, 并给出了q-Bernstein-Durrmeyer型算子在紧圆盘对解析函数的等价定理. 结果表明q-Bernstein-Durrmeyer型算子从实空间推广到复空间扩展了逼近性质.
关键词q-Bernstein-Durrmeyer算子    等价定理    Voronovskaja型定理    
EQUIVALENCE OF COMPLEX APPROXIMATION BY q-BERSTEIN-DURRMEYER OPERATOR IN COMPACT DISKS($0<q<1$)
JIA Yi-xin, HAN Ling-xiong    
School of Mathematical Sciences, Inner Mongolia Minzu University, Tongliao 028000, China
Abstract: This paper studies the relevant properties of the complex q-Bernstein-Durrmeyer type operators in the compact disk. By using the higher-order Cauchy integral formula, Taylor expansion and Bernstein inequality, the simultaneous approximation of the operator in the compact disk and the Voronovskaja-type theorem on the closed unit disk were obtained, and the equivalence theorem of the q-Bernstein-Durrmeyer type operator for analytic functions in the compact disk was given. The results show that the q-Bernstein-Durrmeyer type operator extends the approximation property from real space to complex space.
Keywords: q-Bernstein-Durrmeyer operator     Equivalence theorem     Voronovskaja type results    
1 预备知识与主要结果

本文要用到以下符号, 见文[1], 对$ \forall k>0 $

$ [k]_{q}:=\left\{ \begin{aligned} & \frac{1-q^{k}}{1-q}, &q>0, q\neq 1, \\ &k, & q=1. \end{aligned} \right. $

$ \forall k\in \mathbb{N} $

$ [k]_{q}!:=[1]_{q}[2]_{q}\cdots[k]_{q}, \; [0]!=1. $

对整数$ 0\leq k\leq n, $ $ q $-二项式系数定义为

$ {n\choose k}_{q}:=\frac{[n]_{q}!}{[k]_{q}![n-k]_{q}!}. $

对固定$ q>1 $, $ q $-导数定义为

$ D_{q}f(z)=\left\{ \begin{aligned} & \frac{f(qz)-f(z)}{(q-1)z}, &z\neq 0, \\ &f'(0), & z=0. \end{aligned} \right. $

$ |q|>1 $$ 0<|q|<1 $$ |z|<\frac{1}{1-q} $, 则$ q $-指数函数定义为

$ e_{q}(z):=\sum\limits_{k=0}^{\infty}\frac{z^{k}}{[k]_{q}!}. $

$ |q|>1 $时, 有

$ e_{q}(z)=\prod\limits_{j=0}^{\infty}\biggl(1+(q-1)\frac{z}{q^{j+1}}\biggr), |q|>1. $

算子在紧圆盘的复逼近相关研究已有很多, 如文[2-4]分别研究了复Baskakov-Stancu算子、复Szász-Durrmeyer算子和复Bernstein-Schurer算子在紧圆盘上对解析函数的逼近性质, 但关于$ q $-算子在紧圆盘上逼近性质的相关研究较少. 1997年Philips[5]首次引入并研究了$ q $-Bernstein算子, 之后$ q $-算子的逼近理论成为算子逼近理论的主要研究内容之一. Gupta和Wang[6]引入并研究了$ 0<q<1 $时的$ q $-Durrmeyer算子, 而Agarwal和Gupta[7]将算子扩展到了复空间, 研究了Voronovskaja型结论, 并给出复算子在紧圆盘上对解析函数的精确估计.

复Bernstein-Durrmeyer算子[8]定义为

$ U_{n}^\rho(f;z):=\sum\limits_{k=0}^{n}F_{n, k}^\rho(f)P_{n, k}(z):=\sum\limits_{k=1}^{n-1}\bigg(\displaystyle{\int}_{0}^{1}f(t)\mu_{n, k}^\rho(t)\mathrm{d}t\bigg)P_{n, k}(z)+f(0)(1-z)^n+f(1)z^n, $

其中

$ P_{n, k}(z)={n\choose k}z^k(1-z)^{n-k}, 0\leq k\leq n, k, n\in\mathbb{N}, $
$ \mu_{n, k}^\rho(t):=\frac{t^{k\rho-1}(1-t)^{(n-k)\rho-1}}{B(k\rho, (n-k)\rho)}, 1\leq k\leq n-1, $
$ B(x, y)=\displaystyle{\int}_{0}^{1}t^{x-1}(1-t)^{y-1}\mathrm{d}t, x, y>0. $

文[8]研究了上述算子在紧圆盘上的同时逼近, Voronovskaja型结论和等价定理, 同时文[9]已对$ q $-Bernstein-Durrmeyer算子的Voronovskaja型结论进行了初步研究. 本文主要借鉴文[8]和[10]的研究方法及研究思路, 对$ q $-Bernstein-Durrmeyer算子在紧圆盘上的复逼近性质做进一步研究, 文[11]已给出$ q $-Bernstein-Durrmeyer算子在复空间的定义.

$ f:\overline{D_{R}}\rightarrow \mathbb{C} $且在$ {D_{R}} $内解析,

$ M_{n, q}^\rho(f;z)=f(0)P_{n, 0}(q;z)+f(1)P_{n, n}(q;z)+\sum\limits_{k=1}^{n-1}P_{n, k}(q;z)\displaystyle{\int}_{0}^{1}\mu_{n, k(t)}^\rho f(t)\mathrm{d}t , $

其中$ {D_{R}}=\{z\in\mathbb{C}: |z|<R\}, z\in\mathbb{C}, 0<q<1, \rho>0, n\in\mathbb{N}. $

$ P_{n, k}(q;z)={n\choose k}_{q}z^k(1-z)^{n-k}_{q}, $
$ \mu_{n, k}^\rho(t)=\frac{t^{\rho[k]_{q}-1}(1-t)^{\rho([n]_{q}-[k]_{q})-1}}{B(\rho[k]_{q}, \rho([n]_{q}-[k]_{q}))}, 1\leq k\leq n-1, $
$ B(x, y)=\displaystyle{\int}_{0}^{1}t^{x-1}(1-t)^{y-1}\mathrm{d}t, (x, y>0), $

$ B(x, y) $是Euler Beta函数.

$ q=1 $时, Bernstein-Durrmeyer算子等于$ q $-Bernstein-Durrmeyer算子, 但是为了研究$ q $-Bernstein-Durrmeyer算子在紧圆盘上的复逼近相关性质, 本文限制$ 0<q<1. $

$ e_{m}(z)=z^m, M_{n, q}^\rho(f;z)=\sum_{k=0}^{n}F_{n, k}^\rho(f)P_{n, k}(q;z), $其中

$ F_{n, k}^\rho(f)=\left\{ \begin{aligned} &f(0), & k=0, \\ &\displaystyle{\int}_{0}^{1}\mu_{n, k}^\rho(t)f(t)\mathrm{d}t, & 1\leq k\leq n-1, \\ &f(1), & k=n. \end{aligned} \right. $

$ H(D_{R}) $$ D_{R} $上所有解析函数构成集合, 由$ f\in H(D_{R}) $可知

$ f(z)=\sum\limits_{m=0}^{\infty}a_{m}z^m. $

主要结果如下. 下面定理给出了$ M_{n, q}^\rho(f;z) $在单位圆盘上的Voronovskaja型结论.

定理1.1    设$ f:\overline{D_{R}}\rightarrow \mathbb{C} $, 且$ f\in H(D_{R}), 1<R, \rho>0, 0<q<1, $对所有的$ |z|\leq 1 $, $ n\in\mathbb N, $ $ m\in\mathbb N^*, $

$ \begin{aligned} &\bigg|M_{n, q}^\rho(f;z)-f(z)-\frac{(\rho+1)z(1-z)}{2(1+\rho[n]_{q})}f''(z)\bigg|\\ &\leq\frac{(\rho+1)(\rho+3)|z(1-z)|}{(1+\rho[n]_{q})^{2}}\sum\limits_{m=3}^{\infty}|a_{m}|(m-1)^2(m-2)\bigg(m+\frac{2}{1-q}\bigg). \end{aligned} $

定理1.2    设$ 0<q<1, R>1, D_{R}=\big\{z\in\mathbb{C}:|z|<R\big\}, f:D_{R}\rightarrow \mathbb{C} $$ \rho>0 $, 若$ f $不是一个次数小于等于1的多项式, 那么对任意$ r\in [1, R) $, 有

$ \big\|M_{n, q}^\rho(f)-f\big\|_{r}\geq\frac{\rho+1}{2(1+\rho[n]_{q})}C_{r, \rho}(f), n\in\mathbb N, $

其中$ \|f\|_{r}=\max\big\{|f(z)|:|z|\leq r\big\} $, 常数$ C_{r, \rho}(f) $依赖于$ f, r, \rho, $但与$ n $无关.

下面给出算子在复空间的等价定理.

定理1.3    设$ 0<q<1, f:D_{R}\rightarrow \mathbb{C}, $使$ R>1, D_{R}=\big\{z\in\mathbb{C}:|z|<R\big\}, $$ \rho>0 $, 若$ f $不是一个次数小于等于1的多项式, 那么对任意$ r\in [1, R) $, 有

$ \|M_{n, q}^{\rho}(f)-f\|_{r}\sim\frac{\rho+1}{1+\rho[n]_{q}}, n\in\mathbb N, $

其中$ \|f\|_{r}=\max\big\{|f(z)|;|z|\leq r\big\}. $

定理1.4    设$ 0<q<1, D_{R}=\big\{z\in\mathbb{C}:|z|<R\big\}, f:D_{R}\rightarrow \mathbb{C}, $$ D_{R} $中是解析的, 若$ f $不是一个次数为1的多项式, 那么对任意$ r\in [1, R) $, 有

$ \big\|M_{n, q}^{\rho(p)}(f)-f^{(p)}\big\|_{r}\sim\frac{\rho+1}{1+\rho[n]_{q}}, n\in\mathbb N, $

其中等价中的常数取决于$ f, r, r_{1}, p, $但与$ n $无关.

2 相关引理

为了证明上述定理, 先给出一些引理.

引理2.1[9]    设$ f:\overline{D_{R}}\rightarrow \mathbb{C} $, 且$ f\in H(D_{R}), 1<R, 1<r<R, \rho>0, 0<q<1, |z|\leq r, $ $ n\in\mathbb{N}, m\in\mathbb{N^{*}} $

$ \bigg|M_{n, q}^\rho(f;z)-f(z)-\frac{(\rho+1)z(1-z)}{2(1+\rho[n]_{q})}f''(z)\bigg|\leq\frac{(\rho+1)M_{r}(f)}{(1+\rho[n]_{q})^{2}}, $

其中$ M_{r}(f)=\frac{1}{2}r^{2}(1+r)^{2}\sum_{m=3}^{\infty}|a_{m}|\bigg\{(m-1)^3\bigg[m(3\rho+1)+\bigg(\frac{4}{1-q}-6\bigg)\rho\bigg]\bigg\}r^{m-4}. $

引理2.2[12]    解析函数$ f(z) $的导数仍为解析函数, 它的$ n $阶导数为

$ f^{(n)}(z_{0})=\frac{n!}{2\pi i}\displaystyle{\int}_{C}^{}\frac{f(z)}{(z-z_{0})^{n+1}}\mathrm{d}z, $

其中$ C $$ f(z) $解析区域$ D_{R} $内围绕$ z_{0} $的任何一条正向简单曲线, 且它的内部全含$ D_{R} $.

引理2.3[9]    设$ f:\overline{D_{R}}\rightarrow \mathbb{C} $, 且$ f\in H(D_{R}), 1<r<R, \rho>0, 0<q<1, |z|\leq r, $$ n\in\mathbb{N}, m\in\mathbb{N^{*}} $

$ |M_{n, q}^\rho(f;z)-f(z)|\leq\frac{(\rho+1)r(1+r)}{2(1+\rho[n]_{q})}\sum\limits_{m=2}^{\infty}|a_{m}|m(m-1)r^{m-2}. $

引理2.4(同时逼近) 设$ f:\overline{D_{R}}\rightarrow \mathbb{C} $, 且$ f\in H(D_{R}) $, 对任意固定$ 1\leq r<r_{1}<R, \rho>0, 0<q<1, |z|\leq r, m\in\mathbb{N^{*}}, n, p\in\mathbb{N} $, 有

$ |M_{n, q}^{\rho(p)}(f;z)-f^{(p)}(z)|\leq\frac{(\rho+1)C_{r_{1}}(f)}{1+\rho[n]_{q}}\frac{p!r_{1}}{(r_{1}-r)^{p+1}}, $

其中$ C_{r_{1}}(f)=\frac{1}{2}r(1+r)\sum_{m=2}^{\infty}|a_{m}|m(m-1)r^{m-2}. $

     定义$ T $是圆心为$ O $, 半径为$ r_{1}>r $的圆, 对所有$ |z|\leq r $$ v\in T $$ |v-z|\geq r_{1}-r, $由引理2.3, 对于任意的$ |z|\leq r, n\in\mathbb{N}, m\in\mathbb{N^{*}} $

$ \begin{aligned} \big|M_{n, q}^{\rho(p)}(f;z)-f^{(p)}(z)\big|&\leq\frac{p!}{2\pi}\bigg|\displaystyle{\int}_{T}^{}\frac{M_{n, q}^{\rho}(f;v)-f(v)}{(v-z)^{p+1}}\mathrm{d}v\bigg|\\ &\leq\frac{(\rho+1)r(1+r)}{2(1+\rho[n]_{q})}\sum\limits_{m=2}^{\infty}|a_{m}|m(m-1)r^{m-2}\frac{p!}{2\pi}\frac{2\pi r_{1}}{(r_{1}-r)^{p+1}}\\ &:=\frac{(\rho+1)C_{r_{1}}(f)}{1+\rho[n]_{q}}\frac{p!r_{1}}{(r_{1}-r)^{p+1}}, \end{aligned} $

其中

$ C_{r_{1}}(f)=\frac{1}{2}r(1+r)\sum\limits_{m=2}^{\infty}|a_{m}|m(m-1)r^{m-2}. $
3 定理的证明

定理1.1的证明

$ E_{n, m}(z)=M_{n, q}^\rho(e_{m};z)-z^m-\frac{(\rho+1)m(m-1)}{2(1+\rho[n]_{q})}z^{m-1}(1-z), $其中$ E_{n, m}(z) $是次数小于等于$ m $的多项式.

由参考文献[9]可得等式

$ \begin{aligned} M_{n, q}^\rho(e_{m};z)-z^m&=\frac{\rho z(1-z)}{m-1+\rho[n]_{q}}D_{q}(M_{n, q}^\rho(e_{m-1};z))\\ &\quad+\frac{m-1+\rho[n]_{q}z}{m-1+\rho[n]_{q}}(M_{n, q}^\rho(e_{m-1};z)-z^{m-1}) +\frac{m-1}{m-1+\rho[n]_{q}}(1-z)z^{m-1}, \end{aligned} $

又因

$ \begin{aligned} D_{q}E_{n, m-1}(z)&=D_{q}(M_{n, q}^\rho(e_{m-1};z))-[m-1]_{q}z^{m-2}-\frac{(\rho+1)(m-1)(m-2)[m-2]_{q}}{2(1+\rho[n]_{q})}z^{m-3}\\ &\quad+\frac{(\rho+1)(m-1)(m-2)[m-1]_{q}}{2(1+\rho[n]_{q})}z^{m-2}, \end{aligned} $

$ \begin{aligned} &E_{n, m}(z)\\ &=\frac{\rho z(1-z)}{m-1+\rho[n]_{q}}D_{q}\big(M_{n, q}^\rho(e_{m-1};z)\big)+\frac{m-1+\rho[n]_{q}z}{m-1+\rho[n]_{q}}\big(M_{n, q}^\rho(e_{m-1};z)-z^{m-1}\big)\\ &\quad+\frac{m-1}{m-1+\rho[n]_{q}}(1-z)z^{m-1}-\frac{(\rho+1)m(m-1)}{2(1+\rho[n]_{q})}z^{m-1}(1-z)\\ &=\frac{\rho z(1-z)}{m-1+\rho[n]_{q}}D_{q}E_{n, m-1}(z)+\frac{\rho z(1-z)}{m-1+\rho[n]_{q}}\bigg[[m-1]_{q}z^{m-2}+\frac{(\rho+1)(m-1)(m-2)[m-2]_{q}}{2(1+\rho[n]_{q})}z^{m-3}\\ &\quad-\frac{(\rho+1)(m-1)(m-2)[m-1]_{q}}{2(1+\rho[n]_{q})}z^{m-2}\bigg]\\ &\quad+\frac{m-1+\rho[n]_{q}z}{m-1+\rho[n]_{q}}E_{n, m-1}(z)+\frac{m-1+\rho[n]_{q}z}{m-1+\rho[n]_{q}}\frac{(\rho+1)(m-1)(m-2)}{2(1+\rho[n]_{q})}z^{m-2}(1-z)\\ &\quad+\frac{m-1}{m-1+\rho[n]_{q}}(1-z)z^{m-1}-\frac{(\rho+1)m(m-1)}{2(1+\rho[n]_{q})}(1-z)z^{m-1}\\ &=\frac{\rho z(1-z)}{m-1+\rho[n]_{q}}D_{q}E_{n, m-1}(z)+\frac{m-1+\rho[n]_{q}z}{m-1+\rho[n]_{q}}E_{n, m-1}(z)\\ &\quad+\frac{(1-z)z^{m-2}}{2(1+\rho[n]_{q})(m-1+\rho[n]_{q})}\big\{2(1+\rho[n]_{q})\rho z[m-1]_{q}+\rho(1+\rho)(m-1)(m-2)[m-2]_{q}\\ &\quad-\rho z(1+\rho)(m-1)(m-2)[m-1]_{q}+(m-1+\rho[n]_{q}z)(\rho+1)(m-1)(m-2)\\ &\quad+2(1+\rho[n]_{q})(m-1)z-(m-1+\rho[n]_{q})(\rho+1)m(m-1)z\big\}.\\ \end{aligned} $

从而得

$ E_{n, m}(z)=\frac{\rho z(1-z)}{m-1+\rho[n]_{q}}D_{q}E_{n, m-1}(z)+\frac{m-1+\rho[n]_{q}z}{m-1+\rho[n]_{q}}E_{n, m-1}(z)+R_{n, m}(z), $ (1)

其中

$ \begin{aligned} R_{n, m}(z)&=\frac{(1-z)z^{m-2}}{2(1+\rho[n]_{q})(m-1+\rho[n]_{q})}\big\{(\rho+1)(m-1)(m-2)\big[\rho[m-2]_{q}+(m-1)\big]\\ &\quad-z\big[\rho[m-1]_{q}[(\rho+1)(m-1)(m-2)-2]+(m-1)[(\rho+1)(m-1)m-2]\\ &\quad+2\rho^2[n]_{q}(m-1-[m-1]_{q})\big]\big\}. \end{aligned} $

由公式(1)且$ m\geq 2 $可得

$ \begin{aligned} |E_{n, m}(z)|&\leq\frac{|z(1-z)|}{m-1+\rho[n]_{q}}|D_{q}E_{n, m-1}(z)|+|E_{n, m-1}(z)|+|R_{n, m}(z)|\\ &\leq\frac{|z(1-z)|}{1+\rho[n]_{q}}|D_{q}E_{n, m-1}(z)|+|E_{n, m-1}(z)|+|R_{n, m}(z)|. \end{aligned} $

现研究$ m\geq 2 $$ |D_{q}E_{n, m-1}(z)| $的情况, 由于$ E_{n, m-1}(z) $是次数小于等于$ m-1 $的多项式, 则有

$ \begin{aligned} &|D_{q}E_{n, m-1}(z)|\\ \leq&(m-1)||E_{n, m-1}(z)||_{1}\\ \leq&(m-1)\bigg(||M_{n, q}^\rho(e_{m-1};z)-e_{m-1}(z)||_{1}+\bigg\|\frac{(\rho+1)(m-1)(m-2)}{2(1+\rho[n]_{q})}e_{m-2}(1-e_{1})\bigg\|_{1}\bigg)\\ \leq&(m-1)\bigg(\frac{2(\rho+1)}{2(1+\rho[n]_{q})}m(m-1)+\frac{2(\rho+1)(m-1)(m-2)}{2(1+\rho[n]_{q})}\bigg)\\ =&\frac{(\rho+1)(m-1)^{2}m+(\rho+1)(m-1)^{2}(m-2)}{1+\rho[n]_{q}}\\ \leq&\frac{2(\rho+1)(m-1)^{2}m}{1+\rho[n]_{q}}.\\ \end{aligned} $

又因$ 0<q<1, m\geq 2 $时, 有$ [m-1]_{q}\leq m-1, $且由已知$ |z|\leq1 $, 得

$ \begin{aligned} &|R_{n, m}(z)|\\ \leq&\frac{|z(1-z)|}{2(1+\rho[n]_{q})(m-1+\rho[n]_{q})}\big\{(\rho+1)(m-1)(m-2)\big[\rho(m-2)+(m-1)\big]\\ &\quad+\rho(m-1)\big[(\rho+1)(m-1)(m-2)-2\big]+(m-1)\big[(\rho+1)(m-1)m-2\big]\\ &\quad+2\rho^2[n]_{q}(m-1-(m-1))\big\}\\ \leq&\frac{|z(1-z)|}{2(1+\rho[n]_{q})^2}\bigg\{(\rho+1)(m-1)(m-1)\big[\rho(m-2)+(m-1)\big]\\ &\quad+\rho(m-1)\big[(\rho+1)(m-1)(m-2)\big]+(m-1)\big[(\rho+1)(m-1)m\big]+2\rho^2\frac{1}{1-q}2(m-1)\bigg\}\\ \leq&\frac{|z(1-z)|}{2(1+\rho[n]_{q})^2}\bigg\{(\rho+1)(m-1)^2\bigg[2\rho(m-2)+(m-1)+m+\frac{4\rho}{1-q}\bigg]\bigg\}\\ \leq&\frac{|z(1-z)|}{(1+\rho[n]_{q})^2}(\rho+1)(m-1)^2\bigg[\rho\bigg(m-2+\frac{2}{1-q}\bigg)+m\bigg].\\ \end{aligned} $

最后有

$ \begin{aligned} |E_{n, m}(z)|&\leq|E_{n, m-1}(z)|+\frac{|z(1-z)|}{1+\rho[n]_{q}}\frac{2(\rho+1)(m-1)^{2}m}{1+\rho[n]_{q}}\\ &\quad+\frac{|z(1-z)|}{(1+\rho[n]_{q})^2}(\rho+1)(m-1)^2\bigg[\rho\bigg(m-2+\frac{2}{1-q}\bigg)+m\bigg]\\ &=|E_{n, m-1}(z)|+\frac{|z(1-z)|}{(1+\rho[n]_{q})^2}(\rho+1)(m-1)^{2}\bigg[3m+\rho(m-2)+\frac{2\rho}{1-q}\bigg]\\ &\leq|E_{n, m-1}(z)|+\frac{|z(1-z)|}{(1+\rho[n]_{q})^2}(\rho+1)(m-1)^{2}\bigg[m(\rho+3)+\frac{2\rho}{1-q}\bigg]\\ &\leq|E_{n, m-1}(z)|+\frac{|z(1-z)|}{(1+\rho[n]_{q})^2}(\rho+1)(m-1)^{2}(\rho+3)\bigg(m+\frac{2}{1-q}\bigg), \\ \end{aligned} $

又因$ E_{n, 0}(z)=E_{n, 1}(z)=E_{n, 2}(z)=0, $则由$ m=3, 4, 5, \cdots $逐步推导可得

$ \begin{aligned} |E_{n, m}(z)|&\leq\frac{|z(1-z)|(\rho+1)(\rho+3)}{(1+\rho[n]_{q})^2}\sum\limits_{j=3}^{m}(j-1)^{2}\bigg(j+\frac{2}{1-q}\bigg)\\ &\leq\frac{|z(1-z)|(\rho+1)(\rho+3)}{(1+\rho[n]_{q})^2}(m-1)^{2}\bigg(m+\frac{2}{1-q}\bigg)(m-2), \\ \end{aligned} $

$ \begin{aligned} &\bigg|M_{n, q}^\rho(f;z)-f(z)-\frac{(\rho+1)z(1-z)}{2(1+\rho[n]_{q})}f''(z)\bigg|\\ \leq&\sum\limits_{m=3}^{\infty}|a_{m}||E_{n, m}(z)|\\ \leq&\frac{(\rho+1)(\rho+3)|z(1-z)|}{(1+\rho[n]_{q})^{2}}\sum\limits_{m=3}^{\infty}|a_{m}|(m-1)^2(m-2)\bigg(m+\frac{2}{1-q}\bigg).\\ \end{aligned} $

定理1.2的证明

对所有$ z\in D_{R}, n\in\mathbb N $, 有

$ \begin{aligned} &M_{n, q}^\rho(f;z)-f(z)\\ =&\frac{\rho+1}{2(1+\rho[n]_{q})}\bigg\{z(1-z)f''(z)+\frac{2(1+\rho[n]_{q})}{\rho+1}\bigg(M_{n, q}^\rho(f;z)-f(z)-\frac{(\rho+1)z(1-z)}{2(1+\rho[n]_{q})}f''(z)\bigg)\bigg\}, \end{aligned} $

$ \|F+G\|_{r}\geq\big|\|F\|_{r}-\|G\|_{r}\big|\geq\|F\|_{r}-\|G\|_{r} $可得

$ \begin{aligned} &\|M_{n, q}^\rho(f)-f\|_{r}\\ \geq&\frac{\rho+1}{2(1+\rho[n]_{q})}\bigg\{\|e_{1}(1-e_{1})f''\|_{r}-\frac{2(1+\rho[n]_{q})}{\rho+1}\bigg\|M_{n, q}^\rho(f)-f-\frac{(\rho+1)e_{1}(1-e_{1})}{2(1+\rho[n]_{q})}f''\bigg\|_{r}\bigg\}, \end{aligned} $

其中$ \|f\|_{r}=\max\{|f(z)|;|z|\leq r\}. $由假设知, $ f $不是一个在$ D_{R} $上次数小于等于1的多项式, 即$ ||e_{1}(1-e_{1})f''||_{r}>0 $, 若相反, 则对所有$ z\in\overline{D_{R}} $, 有$ z(1-z)f''(z)=0, $显然可以得出$ f''(z)=0 $, 即$ f $是次数小于等于1的多项式, 与假设矛盾.

由引理2.1有

$ \frac{2(1+\rho[n]_{q})}{\rho+1}\bigg|M_{n, q}^\rho(f;z)-f(z)-\frac{(\rho+1)z(1-z)}{2(1+\rho[n]_{q})}f''(z)\bigg|\rightarrow0, n\rightarrow \infty, $

存在$ n_{1} $(仅依赖于$ f $$ r $), 当$ n\geq n_{1} $时有

$ \begin{aligned} \big\|e_{1}(1-e_{1})f''\big\|_{r}-\frac{2(1+\rho[n]_{q})}{\rho+1}\bigg\|M_{n, q}^\rho(f)-f-\frac{(\rho+1)e_{1}(1-e_{1})}{2(1+\rho[n]_{q})}f''\bigg\|_{r}\geq\frac{1}{2}\big\|e_{1}(1-e_{1})f''\big\|_{r}, \end{aligned} $

$ \|M_{n, q}^\rho(f)-f\|_{r}\geq\frac{\rho+1}{2(1+\rho[n]_{q})}\frac{1}{2}\big\|e_{1}(1-e_{1})f''\big\|_{r}. $

$ 1\leq n\leq n_{1}-1 $时, 有

$ \begin{aligned} \|M_{n, q}^\rho(f)-f\|_{r}&\geq\frac{\rho+1}{2(1+\rho[n]_{q})}\bigg(\frac{2(1+\rho[n]_{q})}{\rho+1}\|M_{n, q}(f)-f\|_{r}\bigg)\\ &:=\frac{\rho+1}{2(1+\rho[n]_{q})}T^{\rho}_{r, n, q}(f)\\ &>0.\\ \end{aligned} $

最后有

$ ||M_{n, q}^\rho(f)-f||_{r}\geq\frac{\rho+1}{2(1+\rho[n]_{q})}C_{r, \rho}(f), $

对所有$ n $, 有$ C_{r, \rho}(f)=\min\big\{T^{\rho}_{r, 1, q}(f), \cdots, T^{\rho}_{r, n-1, q}(f), \frac{1}{2}\big\|e_{1}(1-e_{1})f''\big\|_{r}\big\}. $

定理1.3的证明

结合定理1.2和文[9]中的不等式

$ |M_{n, q}^\rho(f, z)-f(z)|\leq\frac{\rho+1}{1+\rho[n]_{q}}C_{r}(f), $

其中, $ C_{r}(f)=\frac{1}{2}r(1+r)\sum_{m=2}^{\infty}|a_{m}|m(m-1)r^{m-2}, $即可得证.

定理1.4的证明

定义$ T $是圆心为$ O $, 半径为$ r_{1}>r\geq1 $的圆, 对所有$ |z|\leq r $$ v\in T $$ |v-z|\geq r_{1}-r. $ 由引理2.4, 对任意$ |z|\leq r, m\in\mathbb{N^{*}}, n, p\in\mathbb{N} $, 有

$ \begin{aligned} ||M_{n, q}^{\rho(p)}(f)-f^{(p)}||_{r}&\leq\frac{(\rho+1)r(1+r)}{2(1+\rho[n]_{q})}\sum\limits_{m=2}^{\infty}|a_{m}|m(m-1)r^{m-2}\frac{p!r_{1}}{(r_{1}-r)^{p+1}}\\ &:=\frac{(\rho+1)p!r_{1}}{(1+\rho[n]_{q})(r_{1}-r)^{p+1}}C_{r_{1}}(f).\\ \end{aligned} $

下面证明$ ||M_{n, q}^{\rho(p)}(f)-f^{(p)}||_{r} $的下限估计, 根据定理1.2的证明, 对所有$ v\in T $$ n, p\in\mathbb{N} $

$ \begin{aligned} &M_{n, q}^{\rho}(f;z)-f(z)\\ =&\frac{\rho+1}{2(1+\rho[n]_{q})}\bigg\{v(1-v)f''(v)+\frac{2(1+\rho[n]_{q})}{\rho+1}\bigg(M_{n, q}^{\rho}(f;v)-f(v)-\frac{(\rho+1)v(1-v)}{2(1+\rho[n]_{q})}f''(v)\bigg)\bigg\}, \end{aligned} $

$ \begin{aligned} M_{n, q}^{\rho(p)}(f;z)-f^{(p)}(z)&=\frac{\rho+1}{2(1+\rho[n]_{q})}\bigg\{\frac{p!}{2\pi i}\displaystyle{\int}_{T}^{}\frac{v(1-v)}{(v-z)^{p+1}}f''(v)\mathrm{d}v\\ &\quad+\frac{2(1+\rho[n]_{q})}{\rho+1}\frac{p!}{2\pi i}\displaystyle{\int}_{T}^{}\frac{M_{n, q}^{\rho}(f;v)-f(v)-\frac{(\rho+1)v(1-v)}{2(1+\rho[n]_{q})}f''(v)}{(v-z)^{p+1}}\mathrm{d}v\bigg\}\\ &=\frac{\rho+1}{2(1+\rho[n]_{q})}\bigg\{\big[z(1-z)f''(z)\big]^{(p)}\\ &\quad+\frac{2(1+\rho[n]_{q})}{\rho+1}\frac{p!}{2\pi i}\displaystyle{\int}_{T}^{}\frac{M_{n, q}^{\rho}(f;v)-f(v)-\frac{(\rho+1)v(1-v)}{2(1+\rho[n]_{q})}f''(v)}{(v-z)^{p+1}}\mathrm{d}v\bigg\}.\\ \end{aligned} $

现对$ ||\cdot||_{r} $进行研究,

$ \begin{aligned} \big\|M_{n, q}^{\rho(p)}(f)-f^{(p)}\big\|_{r}&\geq\frac{1}{2(1+\rho[n]_{q})}\bigg\{\big\|\big[e_{1}(1-e_{1})f''\big]^{(p)}\big\|_{r}\\ &\quad-\frac{2(1+\rho[n]_{q})}{\rho+1}\bigg\|\frac{p!}{2\pi }\displaystyle{\int}_{T}^{}\frac{M_{n, q}^{\rho}(f;v)-f(v)-\frac{(\rho+1)v(1-v)}{2(1+\rho[n]_{q})}f''(v)}{(v-z)^{p+1}}\mathrm{d}v\bigg\|_{r}\bigg\}, \\ \end{aligned} $

由引理2.1得

$ \begin{aligned} &\bigg\|\frac{p!}{2\pi }\displaystyle{\int}_{T}^{}\frac{M_{n, q}^{\rho}(f;v)-f(v)-\frac{(\rho+1)v(1-v)}{2(1+\rho[n]_{q})}f''(v)}{(v-z)^{p+1}}\mathrm{d}v\bigg\|_{r}\\ \leq&\frac{p!}{2\pi }\frac{2\pi r_{1}}{(r_{1}-r)^{p+1}}\bigg\|M_{n, q}^{\rho}(f)-f-\frac{(\rho+1)e_{1}(1-e_{1})}{2(1+\rho[n]_{q})}f''\bigg\|_{r_{1}}\\ \leq&\frac{p!r_{1}M_{r_{1}}(f)}{(r_{1}-r)^{p+1}}.\\ \end{aligned} $

根据$ f $的假设, 有$ ||[e_{1}(1-e_{1})f'']^{(p)}||_{r}>0 $, 反之, $ z(1-z)f''(z) $是次数$ \leq p-1 $的多项式.

$ p=1 $, 则$ z(1-z)f''(z)=C $, 即$ f''(z)=\frac{C}{z(1-z)}(|z|\leq r, r\geq1) $, 但$ f''(z) $$ |z|\leq r $上解析, 所以$ C=0 $, 即$ f $是次数$ \leq1=\max\{1, p-1\} $的多项式, 矛盾.

$ p=2 $, 则$ z(1-z)f''(z)=Az+B $, 即$ f''(z)=\frac{Az+B}{z(1-z)}(|z|\leq r) $, 但$ f''(z) $$ |z|\leq r $上解析, 所以$ A=B=0 $, 即$ f $是次数$ \leq1=\max\{1, p-1\} $的多项式, 矛盾.

$ p\geq3 $, 则$ z(1-z)f''(z)=U_{p-1}(z), U_{p-1}(z) $是一个次数$ \leq p-1 $的多项式, 即$ f''(z)=\frac{U_{p-1}(z)}{z(1-z)}(|z|\leq r) $, 但$ f''(z) $$ |z|\leq r $上解析, 所以$ U_{p-1}(z)=z(1-z)Q_{p-3}(z), Q_{p-3}(z) $是一个次数$ \leq p-3 $的多项式, 因此$ f''(z)=Q_{p-3}(z) $, 即$ f $是次数$ \leq p-1=\max\{1, p-1\} $的多项式, 矛盾.

综上所述, $ ||[e_{1}(1-e_{1})f'']^{(p)}||_{r}>0 $, 按照定理1.2的方法继续推理即可得出结论.

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